A single-phase voltage source v s=325(250t) V delivers a current, i=12(250t)+9(2150t) A to a load. The load power factor, correct up to two decimal places, is:

GATE 2026 EE – Question 26

Electric Circuits · Transient response of DC and AC networks, sinusoidal steady-state analysis, resonance, two port networks, balanced three phase circuits, star-delta transformation, complex power and power factor in AC circuits · 1 mark · Multiple choice

A single-phase voltage source $v_s=325\sin(2\pi50t)$ V delivers a current, $i=12\sin(2\pi50t)+9\sin(2\pi150t)$ A to a load.

The load power factor, correct up to two decimal places, is:

1.00
0.80
0.65
0.57

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Correct answer: (B) 0.80

Explanation

Only the fundamental components contribute to real power: $P=\dfrac{325}{\sqrt2}\cdot\dfrac{12}{\sqrt2}=1950$ W. The rms values are $V=325/\sqrt2=229.8$ V and $I=\sqrt{(12^2+9^2)/2}=10.61$ A, so $S=VI=2437.5$ VA. The power factor is $1950/2437.5=0.80$.